Making solutions

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Making solutions

  1. 1. Making Solutions Biotechnology I
  2. 2. Introduction  Making solutions is important in any area of biotechnology  R&D QC Mfg  Accuracy is critical as an incorrect solution can destroy months or years worth of work and could delay a critical drug from being available to the market place
  3. 3. Key Terms  Solute: substances that are dissolved  Solvent: substances in which solutes are dissolved (often times this is water or a buffer)  Concentration: amount per volume  Solution: a homogeneous mixture in which one or more substances are dissolved in another.
  4. 4. Concentration 5 mL If each star represents 1 mg of NaCl, what is the total amt. of NaCl in the tube? What is the conc. of NaCl in the tube in mg/mL? 4 mg 4 mg / 5mL = 0.8 mg/mL
  5. 5. Calculations  There are a few equations you need to remember in order to determine the required amount of solute and solvent required.  Take your time with your calculations  Remember to cancel units  Ask yourself if the answer makes sense  Record all calculations and formulas used
  6. 6. Weight per Volume  Simplest way of expressing a concentration  Example: 2mg / mL  2mg is the weight of something in 1 mL of solvent
  7. 7. Percents  May be expressed as weight per volume which is grams per 100 mL (w/v%)  Example: 20 grams of KCL in 100 mL of solvent is a 20% solution (w/v)
  8. 8. Practice Problem  How would you prepare 500 mL of a 5% (w/v) solution of NaCl?  5% = 5g/100mL  5 = x 100 500mL 100x = 2500 x=25grams of NaCl  Weigh out 25 grams of NaCl and add it to a 500 mL volumetric flask. Add approximately 400 mL and mix. Once dissolved, QS to 500 mL. Cap and store.
  9. 9. Percents continued  May be expressed in volume percent which is (v/v): mL of solute 100 mL of solvent  May be expressed in weight per weight which is (w/w): grams solute 100 grams solvent
  10. 10. Practice Problem  How would you make 500 grams of a 5% solution of NaCl by weight (w/w)? % strength is 5% w/w, total weight is 500 grams 5 grams x 500 g = 25 grams of NaCl 100 grams 500 grams total weight – 25 grams NaCl = 475 grams of solvent needed Weigh out 25 grams of NaCl and add it to 475 grams of water.
  11. 11. Molarity  Most common solution calculation used  Definition: molarity (M) is equal to the number of moles of solute that are dissolved per Liter of solvent.  In order to calculate molarity, you need to know moles.  Definition: moles are the molecular weight or formula weight of a particular chemical.  FW or MW can be found on the chemical bottle or by adding up atomic weights of each atom in that molecule.  For example: what is the MW of H20  2x1 + 16 = 18
  12. 12. Practice Calculations  What is the MW for Sulfuric Acid?  H2SO4  2 x 1.00 = 2.00  1 x 32.06= 32.06  4 x 16 = 64.00 98.06  A 1M solution of sulfuric acid contains 98.06 g of sulfuric acid in 1 L of total solution  Mole is an expression of amount  Molarity is an expression of concentration
  13. 13. Key unit conversions  Millimolar (mM)  A millimole is 1/1000 of a mole  Micromolar (µM)  A µmole is 1/1,000,000 of a mole
  14. 14. Practice Problem  Prepare 100 mL of a 0.1M Tris buffer  Step 1: what is the MW of Tris  FW = 121.1 g/mole  What formula are you going to use  Molarity  moles per liter  121.1 grams per 1L or 1000 mL = moles  How much Tris are you going to weight out?
  15. 15. Practice Problem continued  Start with what you know  1M = 121.1 grams/1L  To make 100 mL of 0.1M solution you would add how much Tris buffer?  Formula= FW x molarity x volume (L) = g needed 121.1 x 0.1 x 0.1L = 1.211 grams Tris Weight out 1.211 grams of Tris and add it to 80 mL of water, mix and QS to 100 mL of 0.1M Tris soln.
  16. 16. Practice Problem  How much solute is required to make 400 mL of 0.8 M CaCl2  First what is the FW of CaCl2?  40.08 + (35.45x2) = 110.98  110.98 x 0.8 x 0.4L = 35.51 grams CaCl2 required  Weigh out 35.51 grams CaCl2 and add it to approximately 300 mL of water. Mix and then QS to 400 mL using a graduated cylinder.
  17. 17. Preparing dilute solutions from concentrated solutions  Formula: C1xV1 = C2xV2  C1: concentration of stock solution  V1: volume of stock solution required  C2: concentration you want your dilute solution to be  V2: how much of the dilute solution is required
  18. 18. Practice Problem  How would you prepare 500 mL of a 1M Tris buffer from a 3M stock solution?  What is C1? 3M  What is V1? Unknown  What is C2? 1M  What is V2? 500 mL  3x = 1(500)  X=166.67 mL  Measure 166.67 mL of 3M Tris and add it to 333.33 mL of water to make 500 mL of 1M Tris
  19. 19. Serial Dilutions or Stock Solutions Formula  C1 V1 = C2 V2  Example: Prepare 100 mL of 0.15M KCL from 10 M KCL stock solution (0.15M)(100mL) = (10M)(x mL) 15 = 10x X=1.5 mL You would add 1.5 mL of the 10M KCL solution to 98.5 mL of water for a total volume of 100 mL and a final concentration of 0.15M
  20. 20. Serial Dilutions  In this type of dilution, an aliquot is taken from a stock tube and placed in the first tube. A volume from the second tube is then transferred to the 3rd tube. 1:2 Add 1.0 mL to 1.0 mL 1:4 1:8

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